Proving Geodesic Length Unchanged by Small Curve Changes

[Mmmm]

Homework Statement

Prove that the proper length of geodesic between two points is unchanged to first order by small changes in the curve that do not change its endpoints.

Homework Equations

$$Length of curve = \int ^{\lambda_{2}}_{\lambda_{1}}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}\right|^\frac{1}{2}d\lambda$$

where $g_{\alpha\beta}$ is the metric for the particular coordinate system, and
$\frac{dx^\alpha}{d\lambda}= U^\alpha$ is the gradient of the curve.

On a geodesic, $g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}$ is constant.

The Attempt at a Solution

The equation of the curve is
$$x^\alpha = x^\alpha(\lambda)$$

For a small change in the curve, the equation becomes
$$x^\alpha = x^\alpha(\lambda) + \delta x^\alpha(\lambda)$$
where
$$\delta x^\alpha(\lambda_{2})=\delta x^\alpha(\lambda_{1}) = 0$$
to ensure that the ends of the curve are unchanged.

So

$$Length of curve = \int ^{\lambda_{2}}_{\lambda_{1}}\left|g_{\alpha\beta}\frac{d}{d\lambda}(x^\alpha + \delta x^\alpha)\frac{d}{d\lambda}(x^\alpha + \delta x^\alpha)\right|^\frac{1}{2}d\lambda$$

Multiplying out:
$$=\int ^{\lambda_{2}}_{\lambda_{1}}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}\right|^\frac{1}{2}d\lambda$$

Expanding to first order:

$$\approx \int ^{\lambda_{2}}_{\lambda_{1}}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}\right|^\frac{1}{2}+\frac{1}{2}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}\right|^\frac{-1}{2} \left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}\right|d\lambda$$

The first term here is the length of the original curve again so the second term is the change in length to first order given a small change in the curve. I must prove that this is 0.

$$\Delta l = \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}\right|^\frac{-1}{2} \left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}\right|d\lambda$$

Now on a geodesic $g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}$ is constant, so i will just call this factor C.

so
$$\Delta l = C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|g_{\alpha\beta}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\beta}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\beta}{d\lambda}\right|d\lambda$$

Rename some indices to get a common factor of $\frac{d\delta x^\gamma}{d\lambda}$:

$$= C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|g_{\alpha\gamma}\frac{dx^\alpha}{d\lambda}\frac{d\delta x^\gamma}{d\lambda}+g_{\gamma\beta}\frac{d\delta x^\gamma}{d\lambda}\frac{dx^\beta}{d\lambda}+g_{\alpha\gamma}\frac{d\delta x^\alpha}{d\lambda}\frac{d\delta x^\gamma}{d\lambda}\right|d\lambda$$

Factorise and rename more indices:

$$= C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|\frac{d\delta x^\gamma}{d\lambda}\right|\left|g_{\alpha\gamma}\frac{d x^\alpha}{d\lambda}+g_{\gamma\alpha}\frac{dx^\alpha}{d\lambda}+g_{\alpha\gamma}\frac{d\delta x^\alpha}{d\lambda}\right|d\lambda$$

use symmetry of g to make first two terms equal:

$$= C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|\frac{d\delta x^\gamma}{d\lambda}\right|\left| 2 g_{\alpha\gamma}\frac{d x^\alpha}{d\lambda}+g_{\alpha\gamma}\frac{d\delta x^\alpha}{d\lambda}\right|d\lambda$$

Now integrate by parts:

$$= C \left[ \frac{1}{2}\left|{\delta x^\gamma}\left( 2 g_{\alpha\gamma}\frac{d x^\alpha}{d\lambda}+g_{\alpha\gamma}\frac{d\delta x^\alpha}{d\lambda}\right) \right| \right] ^{\lambda_{2}}_{\lambda_{1}} - C \int ^{\lambda_{2}}_{\lambda_{1}}\frac{1}{2}\left|\delta x^\gamma}\left( 2 \frac{d}{d \lambda} \left( g_{\alpha\gamma} U^\alpha \right) + \frac{d}{d \lambda} \left( g_{\alpha\gamma} \frac{d \delta x^\alpha}{d\lambda} \right) \right)\right|d\lambda$$

The first term vanishes as $\delta x^\alpha(\lambda_{2})=\delta x^\alpha(\lambda_{1}) = 0$

$$= C \int ^{\lambda_{2}}_{\lambda_{1}} \left|\delta x^\gamma}\left[ - \frac{d}{d \lambda} \left( g_{\alpha\gamma} U^\alpha \right) - \frac{1}{2} \frac{d}{d \lambda} \left( g_{\alpha\gamma} \frac{d \delta x^\alpha}{d\lambda} \right) \right]\right|d\lambda$$

This is where I get a bit stuck. The expression in square brackets should come out as the geodesic equation and it is nearly there. The problem is the second term, it should be

$$\frac{1}{2}g_{\alpha \beta , \gamma} U^\alpha U^\beta$$

(which I got from the answer in the back of the book)
to give

$$\Delta l = C \int ^{\lambda_{2}}_{\lambda_{1}} \left|\delta x^\gamma}\left[ - \frac{d}{d \lambda} \left( g_{\alpha\gamma} U^\alpha \right) + \frac{1}{2}g_{\alpha \beta , \gamma} U^\alpha U^\beta \right]\right|d\lambda$$

Then, indeed with a bit of indices tweaking you do get the geodesic equation and everything vanishes.

I have an extra $\frac{d \delta x^\alpha}{d\lambda}$ and I just have no idea how to get rid of it.

 

[Dick]

It's easy to get rid of. It gets rid of itself when the variation goes to zero. It shouldn't have been there to begin with. It came from a term that's second order in the variation. You should have thrown that away at the beginning. The term you are missing comes from the variation of the g, $g(x(\lambda)+\delta x(\lambda))$.

 

[Mmmm]

Aaaaahhhhhhh... of course. after a taylor expansion of g right at the beginning and a removal of that 2nd order term it all comes out beautifully.

Thanks so much Dick... Now I can sleep at night ;)